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Let $(X_n)$ be a sequence of square integrable real random variables on a probability space $(\Omega,\mathcal F,P)$. Suppose that

$$E[X_n\mid \mathcal A]\to 1_A \quad P\text{-a.s.}$$

$$V[X_n\mid \mathcal A]\to 0 \quad P\text{-a.s.}$$

as $n\to \infty$, where $\mathcal A\subset \mathcal F$ is a sub $\sigma$-algebra, and $A\in\mathcal A$ has positive $P$-measure. Can I show then that

$$P[X_n\geq 0]\to 1 \quad \text{as } n\to\infty \quad?$$

Any help is very appreciated.

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  • $\begingroup$ Are you sure that both those conditions can even happen at the same time if $0<P(A)<1$? $\endgroup$
    – Ian
    Jun 17, 2021 at 14:08
  • $\begingroup$ @Ian If $X_n=1_A$ for all $n$ with $A\in\mathcal A$ then I believe both conditions hold no? $\endgroup$
    – Alphie
    Jun 17, 2021 at 14:12
  • $\begingroup$ Never mind, that's right. $\endgroup$
    – Ian
    Jun 17, 2021 at 14:12

2 Answers 2

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No, I don't think we can conclude $P(X_n \ge 0) \rightarrow 1$. Let $\Omega = [0,1]$ with $P$ the Lebesgue measure and $\mathcal A$ the sigma-algebra on $[0,1]$ generated by the Borel subsets of $[0,\frac 12]$. Let $A = [0,\frac 12]$ and \begin{align*} X_n(x) := \begin{cases} 1 & x \in A \\ \frac 1n & x \in (\frac 12, \frac 34] \\ -\frac 1n & x \in (\frac 34,1] \end{cases}. \end{align*} Then $\mathbb{E}[X_n | \mathcal A] = 1_A$ for all $n$ so clearly $\mathbb{E}[X_n | \mathcal A] \rightarrow 1_A$, and $V(X_n | \mathcal A) = \frac{1}{n^2}1_{(\frac 12, 1]} \rightarrow 0$ almost surely. However, we have $P(X_n < 0) = \frac 14$ for all $n$, so $P(X_n \ge 0) \not \rightarrow 1$.

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    $\begingroup$ Typo: the $0$ at the end should be $1$. Anyway, I suppose an obvious followup is "can you ensure $P(X_n>-\epsilon) \to 1$ for each $\epsilon>0$?" $\endgroup$
    – Ian
    Jun 17, 2021 at 16:41
  • $\begingroup$ Isn't $V(X_n | \mathcal A) = 0$ $P$-a.s. for all $n$? $\endgroup$
    – Alphie
    Jun 21, 2021 at 14:03
  • $\begingroup$ @Alphie I think I did make a mistake in $V(X_n|\mathcal A)$, but I don't think it's $0$ a.s. We have $(X_n - \mathbb{E}[X_n|\mathcal A])^2 = (X_n - 1_A)^2 = \frac{1}{n^2} 1_{(\frac 12,1]}$, so $V(X_n|\mathcal A) = \mathbb{E}[(X_n - \mathbb{E}[X_n|\mathcal A])^2|\mathcal A] = \frac{1}{n^2} 1_{(\frac 12,1]}$ which still converges to $0$. $\endgroup$ Jun 21, 2021 at 15:56
  • $\begingroup$ It seems to me that $\mathbb E[1_{(\frac 12,1]} | \mathcal A]=0$ a.s. no? $\endgroup$
    – Alphie
    Jun 21, 2021 at 16:41
  • $\begingroup$ @Alphie No, $(\frac 12,1]$ is measurable with respect to $\mathcal A$ because it is $A^c$, and $A$ is $\mathcal A$-measurable. $\endgroup$ Jun 21, 2021 at 16:49
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In general no: let $X_n=\mathbf{1}_A+Y_n$, where $Y_n$ is independent of $\mathcal A$. The first assumption translates as $\mathbb E\left[Y_n\right]\to 0$ and the second one as $\operatorname{Var}(Y_n)\to 0$.

Moreover, $$ \mathbb P(X_n\geqslant 0)=\mathbb P(A\cap \{1+Y_n\geqslant 0)+\mathbb P(A^c\cap \{Y_n\geqslant 0) $$ and by independence, $$ \mathbb P(X_n\geqslant 0)=\mathbb P(A)\mathbb P(1+Y_n\geqslant 0)+\mathbb P(A^c)\mathbb P(Y_n\geqslant 0). $$ Let $(Y_n)_{n\geqslant 1}$ be a sequence of independent random variables such that for each $n$, $Y_n$ taked the values $1/n$ and $-1/n$ with probability $1/2$. Letting $\mathcal A=\sigma(Y_1)$ and $A=\{Y_1=1\}$, the assumptions on fulfills the assumptions and for $n\geqslant 2$, $$ \mathbb P(X_n\geqslant 0)=\mathbb P(A\cap \{1+X_n\geqslant 0\})+\mathbb P(A^c\cap \{X_n\geqslant 0\})= \mathbb P(A)+\mathbb P(A^c)/2<1, $$ as $\mathbb P(1+Y_n\geqslant 0)=1$ and $\mathbb P(Y_n\geqslant 0)=1/2$.

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  • $\begingroup$ Thank you for your answer. Two questions: 1) Isn't $\mathbb P(X_n\geqslant 0)= \mathbb P(A)+\mathbb P(A^c)/2$? 2) How do you ensure that $Y_n$ is independent from $\mathcal A$? $\endgroup$
    – Alphie
    Jun 21, 2021 at 13:45
  • $\begingroup$ @Alphie Ii have added more details. Actually it seems that $\mathbb P(X_n\geqslant 0)= 1/2$. $\endgroup$ Jun 21, 2021 at 15:12
  • $\begingroup$ Thank you it is clearer now. I would say that $\mathbb P(1+Y_n\geqslant 0)=1$ so that $\mathbb P(X_n\geqslant 0)= 3/4$. $\endgroup$
    – Alphie
    Jun 21, 2021 at 15:38
  • $\begingroup$ Oh you are right, I edited again. $\endgroup$ Jun 21, 2021 at 15:41

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